Difference between revisions of "2017 AMC 10A Problems/Problem 25"
Mathguy1492 (talk | contribs) |
Mathguy1492 (talk | contribs) m |
||
Line 1: | Line 1: | ||
+ | Problem[edit] | ||
How many integers between <math>100</math> and <math>999</math>, inclusive, have the property that some permutation of its digits is a multiple of <math>11</math> between <math>100</math> and <math>999</math>? For example, both <math>121</math> and <math>211</math> have this property. | How many integers between <math>100</math> and <math>999</math>, inclusive, have the property that some permutation of its digits is a multiple of <math>11</math> between <math>100</math> and <math>999</math>? For example, both <math>121</math> and <math>211</math> have this property. | ||
<math> \mathrm{(A) \ }226\qquad \mathrm{(B) \ } 243 \qquad \mathrm{(C) \ } 270 \qquad \mathrm{(D) \ }469\qquad \mathrm{(E) \ } 486</math> | <math> \mathrm{(A) \ }226\qquad \mathrm{(B) \ } 243 \qquad \mathrm{(C) \ } 270 \qquad \mathrm{(D) \ }469\qquad \mathrm{(E) \ } 486</math> | ||
+ | |||
+ | Solution[edit] | ||
+ | |||
Let the three-digit number be <math>ACB</math>: | Let the three-digit number be <math>ACB</math>: | ||
Revision as of 17:42, 8 February 2017
Problem[edit]
How many integers between and
, inclusive, have the property that some permutation of its digits is a multiple of
between
and
? For example, both
and
have this property.
Solution[edit]
Let the three-digit number be :
If a number is divisible by 11, then the difference between the sums of alternating digits is a multiple of 11.
There are two cases:
We now proceed to break down the cases.
:
. This has
cases.
:
, this case results in 110, 220, 330...990. There are two ways to arrange the digits in each of those numbers.
:
, this case results in 121, 231,... 891. There are
ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to
cases.
:
, this case results in 242, 352,... 792. There are
ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to
cases.
:
, this case results in 363, 473,...693. There are
ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to
cases.
:
, this case results in 484 and 594. There are
ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to
cases.
:
.
:
, this cases results in 209, 308, ...506. There are
ways to arrange each of those cases. This leads to
cases.
:
, this cases results in 319, 418, ...616. There are
ways to arrange each of those cases, except the last. This leads to
cases.
:
, this cases results in 429, 528, ...617. There are
ways to arrange each of those cases. This leads to
cases.
...
If you continue this counting, you receive cases.
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.